I have a circuit where resistor is parallel to capacitor, which is charged with voltage U. How to compute line integral around closed loop to get the result of Kirchhof second law - $U_{capacitor}+I\cdot R=0$?

For beginning I splited it in two parts, starting from the capacitor positive plate through resistor:
$$\oint \vec{E} \cdot d \vec{l} = \int_{+R-}\vec{E_{R}} \cdot d \vec{l} +
\int_{-C+} \vec{E_C} \cdot d\vec{l}$$
Since intensity of resistor is with the same direction as current density $\vec{j}$:
$$\int_{+R-}\vec{E_{R}} \cdot d \vec{l} = \int_{+R-} \rho \cdot \vec{j} \cdot d \vec{l} = I \cdot R$$
When I do the same for capacitor I gain negative line integral, because the $E_{c}$ is opposite to $d \vec{l}$ when positive charge is moved from negative plate to the positive.

3 Answers
3

Assign potential $V=0$ to node "-", and potential $V_+$ to node "+". $V_+$ could be either positive or negative, but I'll be assuming positive for definiteness. Clearly, if you compute the line integral $\boldsymbol{E}$ around the circuit, you get a result of 0. I'll use the direction of integration in your question to analyze the elements of that integral. (You could take the opposite sense and get the same result.)

Resistor. Your integral across the resistor, from node "+" to node "-", is correct, with result $IR$. If $V_+> 0$, the electric field will be parallel to $d \boldsymbol{l}$ and you find $I>0$, meaning current is flowing into the "+" node of the resistor, through the resistor, and out the "-" node.

Capacitor, integrating from node "-" to node "+". I'll assign the capacitor potential $U_{capacitor} = V_+$, so that the capacitor's "plus" plate (which could have either sign of charge and voltage) is connected to node "+". Now, if $V_+=U_{capacitor}>0$ the electric field in the capacitor will point from the cap's "plus" plate to its "minus" plate, because $\boldsymbol{E} = - \boldsymbol{\nabla} U$ by definition (that is, $\boldsymbol{E}$ points from high $U$ to low $U$). Then, in your integral, the electric field in the capacitor will be anti-parallel to $d \boldsymbol{l}$, and you get a negative result for this contribution: $-U_{capacitor}.

Adding the two pieces, which must sum to 0, you get:

$$ -U_{capacitor} + IR = 0 \text{ , or}\\
U_{capacitor} = IR $$

Finally, by convention one takes $q$ to be the charge on the "plus" plate of the capacitor (here at node "+"), so that $q=C U_{capacitor}$. Then $dq/dt$ represents the charge flowing into the capacitor's "+" node (through the part, and out the "-" node), which is the opposite of the resistor current:

$$ \frac{dq}{dt} = -I $$

Since $q= C U_{capacitor}$, the above loop equation can be re-written as:

Choose the negative plate of the capacitor as ground, $V=0$. The usual convention is that the electric field lines are drawn in the direction that a positive test charge would move if placed in the field. Then $\int_{-C+} d\vec{l}\cdot \vec{E}_C > 0$ if I traverse the circuit from the negative plate to the positive plate of the capacitor.

So after long I found why the equality:
$$ U_{capacitor} + I\cdot R =0$$
Is somehow correct for calculating passed charge and current in time when the differential equation is solved. The point is that the circulation is calculated correct in way:

$$\oint_{R+C} \vec E \, d\vec l=\int_{+C-} \vec E_C\,d \vec l+\int_{-R+}\vec E_R\,d\vec l= [ \begin{array}f \vec E_C &\mbox{is with the same direction as $d\vec l$}\\
\vec E_R &\mbox{since $\vec J = \sigma\,\vec E_R$ but current in chosen direction is opposite }
\end{array} ] =\\ = \frac{q}{C} - I \cdot R$$
Which according to $\nabla \times \vec E =\vec{0}$ is $0$. If the problem is to get charge $q$ on capacitor at time t, the current is rewritten as $I = |\frac{dq}{dt}|$, but since q on capacitor decreases the $\frac{dq}{dt}<0$ so the result:
$$ \frac{q}{C} + \frac{dq}{dt}=0$$
Which becomes my question if the current direction is chosen to charge the capacitor.